Superpositions and time evolution in a box

The walls force ψ(0) = ψ(L) = 0, which quantises k = nπ/L. The resulting set √(2/L) sin(nπx/L) is complete on the interval (0, L): every square-integrable function that vanishes at both ends can be written as Σ cₙψₙ, which is exactly the Fourier sine series you met in mathematics. A free particle has a continuum of k and needs Ψ(x,0) = ∫ φ(k)e^(ikx) dk/√(2π), an integral over an amplitude density. The box replaces that integral with a sum, and replaces φ(k) with a list of numbers c₁, c₂, c₃, and so on. The difference is not cosmetic. It is why the box has levels you can label and count, why an energy measurement has a discrete menu of possible answers rather than a probability density over a continuum, and why the whole of this lesson is bookkeeping over an index n rather than calculus over a variable k.

To find cₘ, multiply Ψ(x,0) = Σₙ cₙψₙ by ψₘ and integrate across the well. Orthonormality, ∫₀^L ψₘψₙ dx = δₘₙ, kills every term but one and leaves cₘ = ∫₀^L ψₘ(x)Ψ(x,0) dx — the overlap of the state with that level. (In general it is ψₘ* under the integral; the box's real sines make the conjugate invisible here, but it will not stay invisible in later units.) Two shortcuts save most of the labour. Parity: about the midpoint x = L/2, ψₙ is even for odd n and odd for even n, so a state symmetric about the centre has cₙ = 0 for every even n before you integrate anything at all. And Σ|cₙ|² = 1 is not an extra condition to impose; it is the statement that Ψ was normalised, so use it as an arithmetic check on the coefficients you just computed.

Born's rule for the box's non-degenerate spectrum reads P(Eₙ) = |cₙ|², and every other energy statistic follows from that list: ⟨H⟩ = Σ|cₙ|²Eₙ and ⟨H²⟩ = Σ|cₙ|²Eₙ². The mean is a weighted average, not an allowed reading, and the gap between the two ideas can be stark. Take Ψ = √(5/8)ψ₁ + √(3/8)ψ₃. The probabilities are 0.625 and 0.375, and ⟨H⟩ = (5 × 1 + 3 × 9)E₁/8 = 4E₁, which is exactly E₂ — yet c₂ = 0, so a measurement never returns E₂ at all; every reading is E₁ or 9E₁. The spread ΔH = √(⟨H²⟩ − ⟨H⟩²) = √(31 − 16) E₁ = 3.87E₁ is a genuine property of the prepared state, not an instrumental uncertainty that a better apparatus could reduce. What |cₙ|² does not record is the sign or phase of cₙ, and that is exactly the information the next two sections put to work.

Because H is time-independent and the ψₙ are its eigenfunctions, the full solution is Ψ(x, t) = Σ cₙψₙ(x) e^{(−iEₙt/ħ)}. Each coefficient becomes cₙ(t) = cₙ e^{(−iEₙt/ħ)}, and since |e^{(−iEₙt/ħ)}| = 1 the modulus is untouched: the energy statistics and ⟨H⟩ are frozen for ever. What is not frozen is the set of phases, which turn at different rates ωₙ = Eₙ/ħ. Only their differences are observable. If the state is a single eigenfunction, the one phase multiplies the whole wavefunction, cancels in |Ψ|², and nothing whatever happens — that is precisely what the word stationary means. Put two levels in and the relative phase advances at ω₂₁ = (E₂ − E₁)/ħ = 3E₁/ħ. That relative phase is the clock the state runs on, and because E₁ ∝ 1/L², halving the width of the box makes the clock tick four times faster.

Square the two-term state with real c₁ and c₂: |Ψ(x, t)|² = c₁²ψ₁² + c₂²ψ₂² + 2c₁c₂ψ₁ψ₂ cos(ω₂₁t). The first two pieces are exactly the statistical mixture — the density of an ensemble in which a fraction c₁² of the copies sit in ψ₁ and c₂² in ψ₂ — and they never move. Everything that moves is the third piece. Feed the whole thing into ⟨x⟩: each ⟨n|x|n⟩ = L/2, so the diagonal terms together give L/2, and only the off-diagonal matrix element carries time, giving ⟨x⟩(t) = L/2 + 2c₁c₂⟨1|x|2⟩cos(ω₂₁t) with ⟨1|x|2⟩ = −16L/(9π²) = −0.180L. Two consequences follow at once. Since 2c₁c₂ ≤ c₁² + c₂² = 1, no ψ₁–ψ₂ state sloshes further than 0.180L either side of the centre; and reversing the sign of c₂ reverses the side the packet starts on, a difference no probability |cₙ|² can see. Change the pair instead and parity can switch the sliding off entirely, since ⟨n|x|m⟩ = 0 for n ≠ m whenever n + m is even — a ψ₁ + ψ₃ state breathes but never slides.

Eₙ = n²E₁ with n² an integer, so at Tᵣₑᵥ = 2πħ/E₁ = 4mL²/(πħ) every phase has turned through a whole number of cycles and the state returns exactly: 11.0 fs for an electron in a 1.00 nm box. A packet built from many levels does spread in between, because the n² spectrum is dispersive, but the dephasing is undone on schedule because every frequency is a whole multiple of E₁/ħ. Halfway there, at Tᵣₑᵥ/2, the phases are (−1)ⁿ and the density is the exact mirror image of the initial one about x = L/2. Exact revival needs every level spacing to be a whole multiple of a single frequency; the hard-wall ladder has that property, but a finite well or an atom does not, and there revivals are only approximate. The same bookkeeping explains why nothing decays. H has no matrix element between different ψₙ, so every |cₙ|² is conserved and no level is preferred. Even a static coupling between two levels would only swap amplitude back and forth between them; an irreversible decay with a linewidth needs a continuum of final states, such as the modes of the radiation field, which this model does not contain.